TITLE: Combinatorial Identity with Connection Coefficients and Falling Factorial $\langle i x\rangle_n$
QUESTION [7 upvotes]: Let $j, k ,n$ be nonnegative integers such that $0 \leq j, k \leq n \leq k +j $. Pick integer $m$ such that $0 \leq m \leq k + j - n$.
Let $\langle x \rangle_m$ denote the falling factorial $x(x-1)\ldots (x-m+1)$.
I've stumbled across the need to prove the following equality:
$$\sum_{i=0}^n \frac{\binom{k}{i}\binom{j}{i}}{\binom{n}{i}}(-1)^i\langle ix\rangle_m = \left\{ \begin{matrix} \frac{k! j!}{n!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$
An equivalent formulation is:
$$\sum_{i=0}^n \frac{(-1)^i\binom{k}{i}\binom{j}{i} \binom{i x}{m}}{\binom{n}{i}}= \left\{ \begin{matrix} \frac{k! j!}{n!m!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$
Any references to potentially related material would be greatly appreciated.
${\scriptsize \textbf{Edited to fix sign}}$

REPLY [10 votes]: It is very probable that what is written below is the simplification of Darij's argument. I use the notation $x^{\underline{n}}=x(x-1)\dots(x-n+1)$ [as in Knuth's books] for the falling factorial, and $[t^n] f(t)$ for the coefficient of $t^n$ in the polynomial $f$.
For a polynomial $f(t)$ of degree at most $j$ we have $$\sum_{i=0}^j (-1)^{j-i}\frac{f(i)}{i!(j-i)!}=[t^j]f(t),$$
this follows from the Lagrange interpolation of $f$ in the points $\{0,1,\dots,j\}$. Apply this to the polynomial $f(t)=(tx)^{\underline m}(n-t)^{\underline {n-k}}$. In RHS we have 0 if $m<k+j-n$ and $(-1)^{n-k}x^m$ if $m=k+j-n$. In LHS we get $$\sum_{i=0}^{\min(j,k)} 
(-1)^{j-i}\frac{(n-i)!(ti)^{\underline m}}{(k-i)!i!(j-i)!}=(-1)^j\frac{n!}{k!j!}\sum_{i=0}^{\min(j,k)} 
(-1)^{i}\frac{\binom{k}i\binom{j}i}{\binom{n}i}(ti)^{\underline m}$$
and we are done.

REPLY [8 votes]: This is correct. Let me prove a more general fact:

Theorem 1. Let $i$, $j$ and $n$ be three nonnegative integers such that
  $i\leq n$ and $j\leq n$. Let $P\in\mathbb{Q}\left[  X\right]  $ be a
  polynomial such that $\deg P\leq i+j-n$. For every $m\in\mathbb{Z}$, let
  $\left[  X^{m}\right]  P$ denote the coefficient of $X^{m}$ in $P$. Then,
  $\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  = \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!}\left[
X^{i+j-n}\right]  P$.

From this fact, the following follows:

Corollary 2. Let $i$, $j$ and $n$ be three nonnegative integers such that
  $i\leq n$ and $j\leq n$. Let $v\in\mathbb{R}$ and $m\in\left\{  0,1,\ldots
,i+j-n\right\}  $. Then, $\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)
^{u}\dbinom{i}{u}\dbinom{j}{u}}{\dbinom{n}{u}}\dbinom{uv}{m}=
\begin{cases}
 \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}, & \text{if }m=i+j-n;\\
0, & \text{if }m<i+j-n
\end{cases}
$.

Corollary 2 (applied to $k$ and $x$ instead of $i$ and $v$) yields your
question.
Proof of Corollary 2 using Theorem 1. Define a polynomial $P\in
\mathbb{Q}\left[  X\right]  $ by $P=\dbinom{vX}{m}$. Then, $\deg P\leq m\leq
i+j-n$ and $\left[  X^{m}\right]  P=\dfrac{v^{m}}{m!}$ (because if we expand
the product in the numerator of $P=\dbinom{vX}{m}=\dfrac{\left(  vX\right)
\left(  vX-1\right)  \cdots\left(  vX-m+1\right)  }{m!}$, then the only term
of degree $m$ in $X$ will be $\left(  vX\right)  \left(  vX\right)
\cdots\left(  vX\right)  $, thus leading to the coefficient $\dfrac{v^{m}}
{m!}$).
Since $m\in\left\{  0,1,\ldots,i+j-n\right\}  $, we are in one of the
following two cases:
Case 1: We have $m=i+j-n$.
Case 2: We have $m<i+j-n$.
Let us first consider Case 1. In this case, we have $m=i+j-n$. Hence,
$i+j-n=m$, so that $\left[  X^{i+j-n}\right]  P=\left[  X^{m}\right]
P=\dfrac{v^{m}}{m!}$. But Theorem 1 yields
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  = \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!}\underbrace{\left[
X^{i+j-n}\right]  P}_{=\dfrac{v^{m}}{m!}}= \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}$
$=
\begin{cases}
 \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}, & \text{if }m=i+j-n;\\
0, & \text{if }m<i+j-n
\end{cases}
$ (since $m=i+j-n$).
Since $P\left(  u\right)  =\dbinom{vu}{m}=\dbinom{uv}{m}$, this rewrites as
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}\dbinom{uv}{m}=
\begin{cases}
 \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}, & \text{if }m=i+j-n;\\
0, & \text{if }m<i+j-n
\end{cases}
$.
Hence, Corollary 2 is proven in Case 1.
Let us now consider Case 2. In this case, we have $m<i+j-n$. Thus, $\deg P\leq
m<i+j-n$, so that $\left[  X^{i+j-n}\right]  P=0$. But Theorem 1 yields
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  = \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!}\underbrace{\left[
X^{i+j-n}\right]  P}_{=0}=0$
$=
\begin{cases}
 \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}, & \text{if }m=i+j-n;\\
0, & \text{if }m<i+j-n
\end{cases}
$ (since $m<i+j-n$).
Since $P\left(  u\right)  =\dbinom{vu}{m}=\dbinom{uv}{m}$, this rewrites as
$\sum_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}{u}
}{\dbinom{n}{u}}\dbinom{uv}{m}=
\begin{cases}
 \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!m!}v^{m}, & \text{if }m=i+j-n;\\
0, & \text{if }m<i+j-n
\end{cases}
$.
Hence, Corollary 2 is proven in Case 2.
We now have proven Corollary 2 in both cases (using Theorem 1). It remains to
prove Theorem 1.
In the following, $\mathbb{N}$ means the set $\left\{  0,1,2,\ldots\right\}
$. We will use the following facts:

Lemma 3. If $m\in\mathbb{Z}$, $a\in\mathbb{N}$ and $i\in\mathbb{N}$ are
  such that $i\geq a$, then $\dbinom{m}{i}\dbinom{i}{a}=\dbinom{m}{a}\dbinom{m-a}{i-a}$.

Lemma 3 is the so-called trinomial revision identity, and proving it is a
simple exercise in formal manipulations. Notice that the right definition of
binomial coefficients to use here is $\dbinom{m}{p}=\dfrac{m\left(
m-1\right)  \cdots\left(  m-p+1\right)  }{p!}$, since this works for every
$m\in\mathbb{Z}$ (not only for $m\geq p$).

Lemma 4. Let $N\in\mathbb{Z}$. Let $\mathcal{P}_{N}$ be the set of all
  polynomials $P\in\mathbb{Q}\left[  X\right]  $ of degree $\leq N$. Then,
  $\mathcal{P}_{N}$ is a subspace of the $\mathbb{Q}$-vector space
  $\mathbb{Q}\left[  X\right]  $, and has basis $\left(  \dbinom{X}{0}
,\dbinom{X}{1},\ldots,\dbinom{X}{N}\right)  $.

Lemma 4 is well-known.

Lemma 5. Let $q\in\mathbb{N}$ and $r\in\mathbb{Z}$ and $s\in\left\{
0,1,\ldots,q\right\}  $. Then,
  $\sum\limits_{u=0}^{q}\left(  -1\right)  ^{u}\dbinom{q}{u}\dbinom{r-u}{s}=
\begin{cases}
1, & \text{if }s=q;\\
0, & \text{if }s<q
\end{cases}
$.

Lemma 5 is again a fairly basic fact. You might know it in this very form, or
recognize it as a particular case of the fact that the $q$-th finite
difference of a polynomial of degree $\leq q$ is an easily-described constant.
(Here, the polynomial is $\dbinom{r-X}{s}$, whose degree is $s\leq q$.)
Probably, inclusion-exclusion yields a combinatorial proof for $r$ high
enough. In the interest of getting some sleep this week, I will leave the
proof to the reader.
Here is a corollary of Lemma 5 that will be useful to us:

Lemma 6. Let $q\in\mathbb{N}$ and $Q\in\mathbb{N}$ and $r\in\mathbb{Z}$
  and $s\in\left\{  0,1,\ldots,q\right\}  $ be such that $Q\geq q$. Then,
  $\sum\limits_{u=0}^{Q}\left(  -1\right)  ^{u}\dbinom{q}{u}\dbinom{r-u}{s}=
\begin{cases}
1, & \text{if }s=q;\\
0, & \text{if }s<q
\end{cases}
$.

Proof of Lemma 6. In the sum $\sum\limits_{u=0}^{Q}\left(  -1\right)
^{u}\dbinom{q}{u}\dbinom{r-u}{s}$, all the addends with $u>q$ are zero
(because the factor $\dbinom{q}{u}$ makes them vanish). Hence, we can remove
them from the sum without changing the value of the sum. We thus obtain
$\sum\limits_{u=0}^{Q}\left(  -1\right)  ^{u}\dbinom{q}{u}\dbinom{r-u}{s}
=\sum\limits_{u=0}^{q}\left(  -1\right)  ^{u}\dbinom{q}{u}\dbinom{r-u}{s}=
\begin{cases}
1, & \text{if }s=q;\\
0, & \text{if }s<q
\end{cases}
$
(by Lemma 5).
Now, we can prove Theorem 1:
Proof of Theorem 1. The situation is symmetric with respect to $i$ and $j$.
Hence, we WLOG assume that $i\geq j$ (otherwise, we can just interchange $i$
and $j$).
For every $N\in\mathbb{Z}$, define $\mathcal{P}_{N}$ as in Lemma 4. Then,
$P\in\mathcal{P}_{i+j-n}$ (since $P$ is a polynomial of degree $\deg P\leq
i+j-n$). But Lemma 4 (applied to $N=i+j-n$) yields that $\mathcal{P}_{i+j-n}$
is a subspace of the $\mathbb{Q}$-vector space $\mathbb{Q}\left[  X\right]  $,
and has basis $\left(  \dbinom{X}{0},\dbinom{X}{1},\ldots,\dbinom{X}
{i+j-n}\right)  $. But the equality that we are trying to prove (namely,
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  = \left(-1\right)^{i+j-n} \dfrac{i!j!}{n!}\left[
X^{i+j-n}\right]  P$) is $\mathbb{Q}$-linear in $P$ (in the sense that both
its sides depend $\mathbb{Q}$-linearly in $P$). Hence, we can WLOG assume that
$P$ belongs to the above-mentioned basis of $\mathcal{P}_{i+j-n}$. Assume
this. Thus, $P=\dbinom{X}{p}$ for some $p\in\left\{  0,1,\ldots,i+j-n\right\}
$. Consider this $p$. Hence,
$\left[  X^{i+j-n}\right]  P=
\begin{cases}
\dfrac{1}{p!}, & \text{if }p=i+j-n;\\
0, & \text{if }p<i+j-n
\end{cases}
$
$=\dfrac{1}{p!}
\begin{cases}
1, & \text{if }p=i+j-n;\\
0, & \text{if }p<i+j-n
\end{cases}
$.
In other words,
(0) $p!\left[  X^{i+j-n}\right]  P=
\begin{cases}
1, & \text{if }p=i+j-n;\\
0, & \text{if }p<i+j-n
\end{cases}
$.
But $\deg P=p$ (since $P=\dbinom{X}{p}$). Hence, $p=\deg P\leq\underbrace{i}
_{\leq n}+j-n\leq n+j-n=j$, so that $j\geq p$.
We have $P\left(  u\right)  =\dbinom{u}{p}$ for every $u\in\mathbb{Z}$ (since
$P=\dbinom{X}{p}$). Hence,
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}\underbrace{P\left(  u\right)  }_{=\dbinom{u}{p}}
=\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}\dbinom{u}{p}$
$=\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}}{\dbinom
{n}{u}}\dbinom{j}{u}\dbinom{u}{p}$
(1) $=\sum\limits_{u=p}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}
}{\dbinom{n}{u}}\dbinom{j}{u}\dbinom{u}{p}$
(here, we have removed all the addends with $u<p$, because the factor
$\dbinom{u}{p}$ makes these addends vanish). Note that the interval $\left\{
p,p+1,\ldots,n\right\}  $ might be empty, in which case the sum on the right
hand side of (1) is empty; but this is okay (as usual, empty sums are $0$).
Fix $u\in\left\{  p,p+1,\ldots,n\right\}  $. Lemma 3 (applied to $j$, $p$ and
$u$ instead of $m$, $a$ and $i$) yields
(2) $\dbinom{j}{u}\dbinom{u}{p}=\dbinom{j}{p}\dbinom{j-p}{u-p}$.
Now, forget that we fixed $u$. We thus have proven (2) for each
$u\in\left\{  p,p+1,\ldots,n\right\}  $. Hence, (1) rewrites as
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  $
$=\sum\limits_{u=p}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}}{\dbinom
{n}{u}}\dbinom{j}{p}\dbinom{j-p}{u-p}$
$=\dbinom{j}{p}\sum\limits_{u=p}^{n}\left(  -1\right)  ^{u}\dfrac{\dbinom
{i}{u}}{\dbinom{n}{u}}\dbinom{j-p}{u-p}$
(3) $=\dbinom{j}{p}\sum\limits_{u=p}^{i}\left(  -1\right)  ^{u}
\dfrac{\dbinom{i}{u}}{\dbinom{n}{u}}\dbinom{j-p}{u-p}$
(here, we have removed all the addends with $u>i$, because the factor
$\dbinom{i}{u}$ makes these addends vanish).
Now, fix $u\in\left\{  p,p+1,\ldots,i\right\}  $. Then, $u\leq i\leq n$.
Hence, $n-u\in\mathbb{N}$ and $i-u\in\left\{  0,1,\ldots,n-u\right\}  $.
Hence, the symmetry of Pascal's triangle (i.e., the fact that every
$N\in\mathbb{N}$ and $M\in\left\{  0,1,\ldots,N\right\}  $ satisfy $\dbinom
{N}{M}=\dbinom{N}{N-M}$) yields
$\dbinom{n-u}{i-u}=\dbinom{n-u}{\left(  n-u\right)  -\left(  i-u\right)
}=\dbinom{n-u}{n-i}$.
But Lemma 3 (applied to $n$ and $u$ instead of $m$ and $a$) yields
$\dbinom{n}{i}\dbinom{i}{u}=\dbinom{n}{u}\underbrace{\dbinom{n-u}{i-u}
}_{=\dbinom{n-u}{n-i}}=\dbinom{n}{u}\dbinom{n-u}{n-i}$.
In other words,
(4) $\dfrac{\dbinom{i}{u}}{\dbinom{n}{u}}=\dfrac{\dbinom{n-u}{n-i}
}{\dbinom{n}{i}}$.
(The denominators here are nonzero since $u\leq i\leq n$.)
Now, forget that we fixed $u$. We thus have proven (4) for each
$u\in\left\{  p,p+1,\ldots,i\right\}  $. Hence, (3) rewrites as
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  $
$=\dbinom{j}{p}\sum\limits_{u=p}^{i}\left(  -1\right)  ^{u}\dfrac{\dbinom
{n-u}{n-i}}{\dbinom{n}{i}}\dbinom{j-p}{u-p}$
$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\sum\limits_{u=p}^{i}\left(  -1\right)
^{u}\dbinom{n-u}{n-i}\dbinom{j-p}{u-p}$
$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\sum\limits_{u=0}^{i-p}\underbrace{\left(
-1\right)  ^{u+p}}_{=\left(  -1\right)  ^{p}\left(  -1\right)  ^{u}
}\underbrace{\dbinom{n-\left(  u+p\right)  }{n-i}}_{=\dbinom{\left(
n-p\right)  -u}{n-i}}\underbrace{\dbinom{j-p}{\left(  u+p\right)  -p}
}_{=\dbinom{j-p}{u}}$
(here, we have substituted $u+p$ for $u$ in the sum)
$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\left(  -1\right)  ^{p}\sum\limits_{u=0}
^{i-p}\left(  -1\right)  ^{u}\dbinom{\left(  n-p\right)  -u}{n-i}\dbinom
{j-p}{u}$
(5) $=\dbinom{j}{p}\dbinom{n}{i}^{-1}\left(  -1\right)  ^{p}
\sum\limits_{u=0}^{i-p}\left(  -1\right)  ^{u}\dbinom{j-p}{u}\dbinom{\left(
n-p\right)  -u}{n-i}$.
Recall that $j\geq p$. Hence, $j-p\in\mathbb{N}$. Furthermore, $\underbrace{i}
_{\geq j}-p\geq\underbrace{j}_{\geq p}-p\geq0$ and thus $i-p\in\mathbb{N}$.
Finally, $n-i\in\left\{  0,1,\ldots,j-p\right\}  $ (since $n-\underbrace{i}
_{\leq n}\geq n-n=0$ and $n-i\leq j-p$ (since
$p\leq i+j-n$)). Hence, we
can apply Lemma 6 to $q=j-p$, $Q=i-p$, $r=n-p$ and $s=n-i$. We thus obtain
$\sum\limits_{u=0}^{i-p}\left(  -1\right)  ^{u}\dbinom{j-p}{u}\dbinom{\left(
n-p\right)  -u}{n-i}=
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
$.
Therefore, (5) rewrites as
$\sum\limits_{u=0}^{n}\dfrac{\left(  -1\right)  ^{u}\dbinom{i}{u}\dbinom{j}
{u}}{\dbinom{n}{u}}P\left(  u\right)  $
$=\underbrace{\dbinom{j}{p}}_{=\dfrac{j!}{p!\left(  j-p\right)  !}
}\underbrace{\dbinom{n}{i}^{-1}}_{=\dfrac{i!\left(  n-i\right)  !}{n!}}\left(
-1\right)  ^{p}
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
$
$=\dfrac{j!}{p!\left(  j-p\right)  !}\cdot\dfrac{i!\left(  n-i\right)  !}
{n!}\cdot\left(  -1\right)  ^{p}
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
$
$=\left(  -1\right)  ^{p}\dfrac{i!j!}{n!p!}\cdot\underbrace{\dfrac{\left(
n-i\right)  !}{\left(  j-p\right)  !}
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
}_{\substack{=
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
\\\text{(since }\dfrac{\left(  n-i\right)  !}{\left(  j-p\right)  !}=1\text{
in the case when }n-i=j-p\text{,}\\\text{whereas in the other case both sides
are }0\text{)}}}$
$=\left(  -1\right)  ^{p}\dfrac{i!j!}{n!p!}\cdot
\begin{cases}
1, & \text{if }n-i=j-p;\\
0, & \text{if }n-i<j-p
\end{cases}
$
$=\left(  -1\right)  ^{p}\dfrac{i!j!}{n!p!}\cdot
\begin{cases}
1, & \text{if }p=i+j-n;\\
0, & \text{if }p<i+j-n
\end{cases}
$
(since $n-i=j-p$ is equivalent to $p=i+j-n$)
$=\left(  -1\right)  ^{i+j-n}\dfrac{i!j!}{n!p!}\cdot\underbrace{
\begin{cases}
1, & \text{if }p=i+j-n;\\
0, & \text{if }p<i+j-n
\end{cases}
}_{\substack{=p!\left[  X^{i+j-n}\right]  P\\\text{(by the equality (0))}}}$
(since $\left(  -1\right)  ^{p}=\left(  -1\right)  ^{i+j-n}$ in the case when
$p=i+j-n$, whereas in the other case both sides are $0$)
$=\left(  -1\right)  ^{i+j-n}\dfrac{i!j!}{n!p!}\cdot p!\left[  X^{i+j-n}
\right]  P=\left(  -1\right)  ^{i+j-n}\dfrac{i!j!}{n!}\left[  X^{i+j-n}
\right]  P$.
Thus, Theorem 1 is proven.